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Unformatted text preview: 43. (a) and (b) Using Eq. 406 and the result of problem 3 in Chapter 39, we ﬁnd
∆E = Ephoton = hc
1240 eV · nm
=
= 2.55 eV .
λ
486.1 nm Referring to Fig. 4016, we see that this must be one of the Balmer series transitions (this fact
could also be found from Fig. 4017). Therefore, nlow = 2, but what precisely is nhigh ?
Ehigh
13.6 eV
−
n2 =
= Elow + ∆E
13.6 eV
−
+ 2.55 eV
22 which yields n = 4. Thus, the transition is from the n = 4 to the n = 2 state. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Photon

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